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39th International Physics Olympiad  Hanoi  Vietnam  2008
Theoretical Problem No. 3 / Solution
Solution
1.
For an altitude change
, the atmospheric pressure change is :
dz
dp
gdz
ρ
= −
(1)
where
is the acceleration of gravity, considered constant,
g
is the specific mass of
air,
which is considered as an ideal gas:
mp
VR
T
μ
==
Put this expression in (1) :
dp
g
dz
pR
T
=−
1.1. If the air temperature is uniform and equals
, then
0
T
0
dp
g
dz
T
After integration, we have :
() ()
0
0
e
g
z
RT
pz
p
−
=
(2)
1.2. If
( ) ( )
0
Tz T
z
Λ
(3)
then
()
0
dp
g
dz
p
RT
z
⎡⎤
−Λ
⎣⎦
(4)
1.2.1. Knowing that :
( )
0
11
0
00
ln
dT
z
dz
Tz
Λ
Λ
∫∫
by integrating both members of (4), we obtain :
( )
( )
0
1
ln
ln
ln
T
z
gg
T
R
T
μμ
⎛⎞
Λ
⎜
⎜
ΛΛ
⎝⎠
0
z
−
⎟
⎟
01
0
g
R
z
p
T
Λ
Λ
⎜
⎜
⎟
⎟
(5)
1
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Theoretical Problem No. 3 / Solution
1.2.2. The free convection occurs if:
( )
()
1
0
z
ρ
>
The ratio of specific masses can be expressed as follows:
1
0
1
00
0
g
R
zp
z
T
z
pT
z
T
μ
−
Λ
⎛⎞
Λ
==
−
⎜⎟
⎝⎠
The last term is larger than unity if its exponent is negative:
10
g
R
−<
Λ
Then :
0 029 9 81
K
0034
831
m
..
.
.
g
R
×
Λ>
=
=
2.
In vertical motion, the pressure of the parcel always equals that of the surrounding air,
the latter depends on the altitude. The parcel temperature
parcel
T
depends on the
pressure.
2.1. We can write:
parcel
parcel
dT
dT
dp
dz
dp
dz
=
p
is simultaneously the pressure of air in the parcel and that of the surrounding air.
Expression for
parcel
dT
dp
By using the equation for adiabatic processes
and equation of state,
we can deduce the equation giving the change of pressure and temperature in a
quasiequilibrium adiabatic process of an air parcel:
const
pV
γ
=
1
parcel
const
Tp
−
=
(6)
2
39th International Physics Olympiad  Hanoi  Vietnam  2008
Theoretical Problem No. 3 / Solution
where
p
V
c
c
γ
=
is the ratio of isobaric and isochoric thermal capacities of air. By
logarithmic differentiation
of
the two
members of
(6),
we have:
parcel
parcel
1
0
dT
dp
Tp
−
+=
Or
parcel
parcel
1
dT
T
dp
p
−
=
(7)
Note:
we can use the first law of thermodynamic to calculate the heat received by the
parcel in an elementary process:
parcel
V
m
dQ
c dT
pdV
μ
=+
, this heat equals zero in an
adiabatic process. Furthermore, using the equation of state for air in the parcel
parcel
m
pV
RT
=
we can derive (6)
Expression for
dp
dz
From (1) we can deduce:
dp
pg
g
dz
RT
ρ
=−
where
is the temperature of the surrounding air.
T
On the basis of these two expressions, we derive the expression for
:
parcel
/
dT
dz
parcel
parcel
1
dT
T
g
G
dz
R
T
γμ
−
(8)
In general,
is not a constant.
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This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.
 Spring '11
 NA
 Acceleration, Gravity

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